NEET Exam > NEET Notes > Physics Class 11 > HC Verma Solutions: Chapter 8 - Work & Energy
HC Verma Solutions: Chapter 8 - Work & Energy
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Page 1
8.1
SOLUTIONS TO CONCEPTS
CHAPTER – 8
1. M = m
c
+ m
b
= 90kg
u = 6 km/h = 1.666 m/sec
v = 12 km/h = 3.333 m/sec
Increase in K.E. = ½ Mv
2
– ½ Mu
2
= ½ 90 × (3.333)
2
– ½ × 90 × (1.66)
2
= 494.5 – 124.6 = 374.8 ? 375 J
2. m
b
= 2 kg.
u = 10 m/sec
a = 3 m/aec
2
t = 5 sec
v = u + at = 10 + 3 I 5 = 25 m/sec.
?F.K.E = ½ mv
2
= ½ × 2 × 625 = 625 J.
3. F = 100 N
S = 4m, ? = 0°
? = S . F
? ?
???100 × 4 = 400 J ?
4. m = 5 kg
? = 30°
S = 10 m
F = mg
So, work done by the force of gravity
? = mgh = 5 × 9.8 × 5 = 245 J ?
5. F= 2.50N, S = 2.5m, m =15g = 0.015kg.
So, w = F × S ? a =
m
F
=
015 . 0
5 . 2
=
3
500
m/s
2
=F × S cos 0° (acting along the same line)
= 2.5 × 2.5 = 6.25J
Let the velocity of the body at b = U. Applying work-energy principle ½ mv
2
– 0 = 6.25
? V =
015 . 0
2 25 . 6 ?
= 28.86 m/sec.
So, time taken to travel from A to B.
? t =
a
u v ?
=
500
3 86 . 28 ?
? Average power =
t
W
=
3 ) 86 . 28 (
500 25 . 6
?
?
= 36.1
6. Given
j
ˆ
3 i
ˆ
2 r
1
? ?
?
j
ˆ
2 i
ˆ
3 r
2
? ?
So, displacement vector is given by,
2 1
r r r
? ? ?
? ? ? j
ˆ
i
ˆ
) j
ˆ
3 i
ˆ
2 ( ) j
ˆ
2 i
ˆ
3 ( r ? ? ? ? ? ?
?
u=1.66 m/s ?
90kg ? 90kg ?
v=3.33 m/s ?
u=10 m/s ?
2 kg ? a
?
= 3m/s
2
F ?
4m ?
R ?
100 N ?
mg ?
30°
?
5 ?
mg ?
F ?
5 log ?
10m ?
30°
?
v ? ?
A ? B ?
Page 2
8.1
SOLUTIONS TO CONCEPTS
CHAPTER – 8
1. M = m
c
+ m
b
= 90kg
u = 6 km/h = 1.666 m/sec
v = 12 km/h = 3.333 m/sec
Increase in K.E. = ½ Mv
2
– ½ Mu
2
= ½ 90 × (3.333)
2
– ½ × 90 × (1.66)
2
= 494.5 – 124.6 = 374.8 ? 375 J
2. m
b
= 2 kg.
u = 10 m/sec
a = 3 m/aec
2
t = 5 sec
v = u + at = 10 + 3 I 5 = 25 m/sec.
?F.K.E = ½ mv
2
= ½ × 2 × 625 = 625 J.
3. F = 100 N
S = 4m, ? = 0°
? = S . F
? ?
???100 × 4 = 400 J ?
4. m = 5 kg
? = 30°
S = 10 m
F = mg
So, work done by the force of gravity
? = mgh = 5 × 9.8 × 5 = 245 J ?
5. F= 2.50N, S = 2.5m, m =15g = 0.015kg.
So, w = F × S ? a =
m
F
=
015 . 0
5 . 2
=
3
500
m/s
2
=F × S cos 0° (acting along the same line)
= 2.5 × 2.5 = 6.25J
Let the velocity of the body at b = U. Applying work-energy principle ½ mv
2
– 0 = 6.25
? V =
015 . 0
2 25 . 6 ?
= 28.86 m/sec.
So, time taken to travel from A to B.
? t =
a
u v ?
=
500
3 86 . 28 ?
? Average power =
t
W
=
3 ) 86 . 28 (
500 25 . 6
?
?
= 36.1
6. Given
j
ˆ
3 i
ˆ
2 r
1
? ?
?
j
ˆ
2 i
ˆ
3 r
2
? ?
So, displacement vector is given by,
2 1
r r r
? ? ?
? ? ? j
ˆ
i
ˆ
) j
ˆ
3 i
ˆ
2 ( ) j
ˆ
2 i
ˆ
3 ( r ? ? ? ? ? ?
?
u=1.66 m/s ?
90kg ? 90kg ?
v=3.33 m/s ?
u=10 m/s ?
2 kg ? a
?
= 3m/s
2
F ?
4m ?
R ?
100 N ?
mg ?
30°
?
5 ?
mg ?
F ?
5 log ?
10m ?
30°
?
v ? ?
A ? B ?
Chapter 8
8.2
So, work done = s F
?
?
? = 5 × 1 + 5(-1) = 0
7. m
b
= 2kg, s = 40m, a = 0.5m/sec
2
So, force applied by the man on the box
F = m
b
a = 2 × (0.5) = 1 N
? = FS = 1 × 40 = 40 J ?
8. Given that F= a + bx
Where a and b are constants.
So, work done by this force during this force during the displacement x = 0 and x = d is given
by
W =
? ?
? ?
d
0
d
0
dx ) bx a ( dx F = ax + (bx
2
/2) = [a + ½ bd] d
9. m
b
= 250g = .250 kg
? = 37°, S = 1m.
Frictional force f = ?R
mg sin ? = ? R ..(1)
mg cos ?? ..(2)
so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J
10. a =
) m M ( 2
F
?
(given)
a) from fig (1)
ma = ?
k
R
1
and R
1
= mg
? ? =
1
R
ma
=
g ) m M ( 2
F
?
b) Frictional force acting on the smaller block f = ?R =
) m M ( 2
F m
mg
g ) m M ( 2
F
?
?
? ?
?
c) Work done w = fs s = d
w = d
) m M ( 2
mF
?
?
=
) m M ( 2
mFd
?
?
11. Weight = 2000 N, S = 20m, ? = 0.2
a) R + Psin ? - 2000 = 0 ..(1)
P cos ? - 0.2 R =0 ..(2)
From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0
P =
? ? ? sin 2 . 0 cos
400
..(3)
So, work done by the person, W = PS cos ? =
? ? ?
?
sin 2 . 0 cos
cos 8000
=
? ? sin 2 . 0 1
8000
=
? ? tan 5
40000
b) For minimum magnitude of force from equn(1)
d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2
putting the value in equn (3)
W =
? ? tan 5
40000
=
) 2 . 5 (
40000
= 7690 J
12. w = 100 N, ? = 37°, s = 2m
R ?
m b g ?
m b a ? F ?
R ?
?R ?
1 m ?
mg ?
37° ?
M
?
F ?
m
?
R 1 ?
? k R 1 ?
ma ?
mg ?
?R 1 ?
R 2 ?
f
?
ma ?
mg ?
?R 2 ?
? ?
R
?
P
?
0.2R ?
2000 N ?
Page 3
8.1
SOLUTIONS TO CONCEPTS
CHAPTER – 8
1. M = m
c
+ m
b
= 90kg
u = 6 km/h = 1.666 m/sec
v = 12 km/h = 3.333 m/sec
Increase in K.E. = ½ Mv
2
– ½ Mu
2
= ½ 90 × (3.333)
2
– ½ × 90 × (1.66)
2
= 494.5 – 124.6 = 374.8 ? 375 J
2. m
b
= 2 kg.
u = 10 m/sec
a = 3 m/aec
2
t = 5 sec
v = u + at = 10 + 3 I 5 = 25 m/sec.
?F.K.E = ½ mv
2
= ½ × 2 × 625 = 625 J.
3. F = 100 N
S = 4m, ? = 0°
? = S . F
? ?
???100 × 4 = 400 J ?
4. m = 5 kg
? = 30°
S = 10 m
F = mg
So, work done by the force of gravity
? = mgh = 5 × 9.8 × 5 = 245 J ?
5. F= 2.50N, S = 2.5m, m =15g = 0.015kg.
So, w = F × S ? a =
m
F
=
015 . 0
5 . 2
=
3
500
m/s
2
=F × S cos 0° (acting along the same line)
= 2.5 × 2.5 = 6.25J
Let the velocity of the body at b = U. Applying work-energy principle ½ mv
2
– 0 = 6.25
? V =
015 . 0
2 25 . 6 ?
= 28.86 m/sec.
So, time taken to travel from A to B.
? t =
a
u v ?
=
500
3 86 . 28 ?
? Average power =
t
W
=
3 ) 86 . 28 (
500 25 . 6
?
?
= 36.1
6. Given
j
ˆ
3 i
ˆ
2 r
1
? ?
?
j
ˆ
2 i
ˆ
3 r
2
? ?
So, displacement vector is given by,
2 1
r r r
? ? ?
? ? ? j
ˆ
i
ˆ
) j
ˆ
3 i
ˆ
2 ( ) j
ˆ
2 i
ˆ
3 ( r ? ? ? ? ? ?
?
u=1.66 m/s ?
90kg ? 90kg ?
v=3.33 m/s ?
u=10 m/s ?
2 kg ? a
?
= 3m/s
2
F ?
4m ?
R ?
100 N ?
mg ?
30°
?
5 ?
mg ?
F ?
5 log ?
10m ?
30°
?
v ? ?
A ? B ?
Chapter 8
8.2
So, work done = s F
?
?
? = 5 × 1 + 5(-1) = 0
7. m
b
= 2kg, s = 40m, a = 0.5m/sec
2
So, force applied by the man on the box
F = m
b
a = 2 × (0.5) = 1 N
? = FS = 1 × 40 = 40 J ?
8. Given that F= a + bx
Where a and b are constants.
So, work done by this force during this force during the displacement x = 0 and x = d is given
by
W =
? ?
? ?
d
0
d
0
dx ) bx a ( dx F = ax + (bx
2
/2) = [a + ½ bd] d
9. m
b
= 250g = .250 kg
? = 37°, S = 1m.
Frictional force f = ?R
mg sin ? = ? R ..(1)
mg cos ?? ..(2)
so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J
10. a =
) m M ( 2
F
?
(given)
a) from fig (1)
ma = ?
k
R
1
and R
1
= mg
? ? =
1
R
ma
=
g ) m M ( 2
F
?
b) Frictional force acting on the smaller block f = ?R =
) m M ( 2
F m
mg
g ) m M ( 2
F
?
?
? ?
?
c) Work done w = fs s = d
w = d
) m M ( 2
mF
?
?
=
) m M ( 2
mFd
?
?
11. Weight = 2000 N, S = 20m, ? = 0.2
a) R + Psin ? - 2000 = 0 ..(1)
P cos ? - 0.2 R =0 ..(2)
From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0
P =
? ? ? sin 2 . 0 cos
400
..(3)
So, work done by the person, W = PS cos ? =
? ? ?
?
sin 2 . 0 cos
cos 8000
=
? ? sin 2 . 0 1
8000
=
? ? tan 5
40000
b) For minimum magnitude of force from equn(1)
d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2
putting the value in equn (3)
W =
? ? tan 5
40000
=
) 2 . 5 (
40000
= 7690 J
12. w = 100 N, ? = 37°, s = 2m
R ?
m b g ?
m b a ? F ?
R ?
?R ?
1 m ?
mg ?
37° ?
M
?
F ?
m
?
R 1 ?
? k R 1 ?
ma ?
mg ?
?R 1 ?
R 2 ?
f
?
ma ?
mg ?
?R 2 ?
? ?
R
?
P
?
0.2R ?
2000 N ?
Chapter 8
8.3
Force F= mg sin 37° = 100 × 0.60 = 60 N
So, work done, when the force is parallel to incline.
w = Fs cos ? = 60 × 2 × cos ? = 120 J
In ?ABC AB= 2m
CB = 37°
so, h = C = 1m
?work done when the force in horizontal direction
W = mgh = 100 × 1.2 = 120 J ?
13. m = 500 kg, s = 25m, u = 72km/h= 20 m/s, v = 0
(-a) =
S 2
u v
2 2
?
? a =
50
400
= 8m/sec
2
Frictional force f = ma = 500 × 8 = 4000 N
14. m = 500 kg, u = 0, v = 72 km/h = 20m/s
a =
s 2
u v
2 2
?
=
50
400
= 8m/sec
2
force needed to accelerate the car F = ma = 500 × 8 = 4000 N
15. Given, v = a x (uniformly accelerated motion)
displacement s = d – 0 = d
putting x = 0, v
1
= 0
putting x = d, v
2
= a d
a =
s 2
u v
2
2
2
2
?
=
d 2
d a
2
=
2
a
2
force f = ma =
2
ma
2
work done w = FS cos ? = d
2
ma
2
? =
2
d ma
2
?
16. a) m = 2kg, ? = 37°, F = 20 N
From the free body diagram
F = (2g sin ?) + ma ? a = (20 – 20 sin ?)/s = 4m/sec
2
S = ut + ½ at
2
(u = 0, t = 1s, a = 1.66)
= 2m
So, work, done w = Fs = 20 × 2 = 40 J
b) If W = 40 J
S =
F
W
=
20
40
h = 2 sin 37° = 1.2 m
So, work done W = –mgh = – 20 × 1.2 = –24 J
c) v = u + at = 4 × 10 = 40 m/sec
So, K.E. = ½ mv
2
= ½ × 2 × 16 = 16 J
17. m = 2kg, ? = 37°, F = 20 N, a = 10 m/sec
2
a) t = 1sec
So, s= ut + ½ at
2
= 5m
37°
?
A ?
A ?
A ?
B ?
v=0
?
v=20 m/s
m=500 kg ?
–a
?
25m
?
a
?
R
mg
?
ma f
?
500 kg ?
a
?
25m
?
R
mg
?
F
F
?
ma
?
ma ?
2g cos ? ?
20N
?
R
?
ma 2gsin ? ?
20N ?
ma ?
mg cos ? ?
20N
?
R
?
mg sin ? ?
?R ?
h
?
37°
?
5m ?
C ?
A ?
B ?
37°
?
Page 4
8.1
SOLUTIONS TO CONCEPTS
CHAPTER – 8
1. M = m
c
+ m
b
= 90kg
u = 6 km/h = 1.666 m/sec
v = 12 km/h = 3.333 m/sec
Increase in K.E. = ½ Mv
2
– ½ Mu
2
= ½ 90 × (3.333)
2
– ½ × 90 × (1.66)
2
= 494.5 – 124.6 = 374.8 ? 375 J
2. m
b
= 2 kg.
u = 10 m/sec
a = 3 m/aec
2
t = 5 sec
v = u + at = 10 + 3 I 5 = 25 m/sec.
?F.K.E = ½ mv
2
= ½ × 2 × 625 = 625 J.
3. F = 100 N
S = 4m, ? = 0°
? = S . F
? ?
???100 × 4 = 400 J ?
4. m = 5 kg
? = 30°
S = 10 m
F = mg
So, work done by the force of gravity
? = mgh = 5 × 9.8 × 5 = 245 J ?
5. F= 2.50N, S = 2.5m, m =15g = 0.015kg.
So, w = F × S ? a =
m
F
=
015 . 0
5 . 2
=
3
500
m/s
2
=F × S cos 0° (acting along the same line)
= 2.5 × 2.5 = 6.25J
Let the velocity of the body at b = U. Applying work-energy principle ½ mv
2
– 0 = 6.25
? V =
015 . 0
2 25 . 6 ?
= 28.86 m/sec.
So, time taken to travel from A to B.
? t =
a
u v ?
=
500
3 86 . 28 ?
? Average power =
t
W
=
3 ) 86 . 28 (
500 25 . 6
?
?
= 36.1
6. Given
j
ˆ
3 i
ˆ
2 r
1
? ?
?
j
ˆ
2 i
ˆ
3 r
2
? ?
So, displacement vector is given by,
2 1
r r r
? ? ?
? ? ? j
ˆ
i
ˆ
) j
ˆ
3 i
ˆ
2 ( ) j
ˆ
2 i
ˆ
3 ( r ? ? ? ? ? ?
?
u=1.66 m/s ?
90kg ? 90kg ?
v=3.33 m/s ?
u=10 m/s ?
2 kg ? a
?
= 3m/s
2
F ?
4m ?
R ?
100 N ?
mg ?
30°
?
5 ?
mg ?
F ?
5 log ?
10m ?
30°
?
v ? ?
A ? B ?
Chapter 8
8.2
So, work done = s F
?
?
? = 5 × 1 + 5(-1) = 0
7. m
b
= 2kg, s = 40m, a = 0.5m/sec
2
So, force applied by the man on the box
F = m
b
a = 2 × (0.5) = 1 N
? = FS = 1 × 40 = 40 J ?
8. Given that F= a + bx
Where a and b are constants.
So, work done by this force during this force during the displacement x = 0 and x = d is given
by
W =
? ?
? ?
d
0
d
0
dx ) bx a ( dx F = ax + (bx
2
/2) = [a + ½ bd] d
9. m
b
= 250g = .250 kg
? = 37°, S = 1m.
Frictional force f = ?R
mg sin ? = ? R ..(1)
mg cos ?? ..(2)
so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J
10. a =
) m M ( 2
F
?
(given)
a) from fig (1)
ma = ?
k
R
1
and R
1
= mg
? ? =
1
R
ma
=
g ) m M ( 2
F
?
b) Frictional force acting on the smaller block f = ?R =
) m M ( 2
F m
mg
g ) m M ( 2
F
?
?
? ?
?
c) Work done w = fs s = d
w = d
) m M ( 2
mF
?
?
=
) m M ( 2
mFd
?
?
11. Weight = 2000 N, S = 20m, ? = 0.2
a) R + Psin ? - 2000 = 0 ..(1)
P cos ? - 0.2 R =0 ..(2)
From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0
P =
? ? ? sin 2 . 0 cos
400
..(3)
So, work done by the person, W = PS cos ? =
? ? ?
?
sin 2 . 0 cos
cos 8000
=
? ? sin 2 . 0 1
8000
=
? ? tan 5
40000
b) For minimum magnitude of force from equn(1)
d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2
putting the value in equn (3)
W =
? ? tan 5
40000
=
) 2 . 5 (
40000
= 7690 J
12. w = 100 N, ? = 37°, s = 2m
R ?
m b g ?
m b a ? F ?
R ?
?R ?
1 m ?
mg ?
37° ?
M
?
F ?
m
?
R 1 ?
? k R 1 ?
ma ?
mg ?
?R 1 ?
R 2 ?
f
?
ma ?
mg ?
?R 2 ?
? ?
R
?
P
?
0.2R ?
2000 N ?
Chapter 8
8.3
Force F= mg sin 37° = 100 × 0.60 = 60 N
So, work done, when the force is parallel to incline.
w = Fs cos ? = 60 × 2 × cos ? = 120 J
In ?ABC AB= 2m
CB = 37°
so, h = C = 1m
?work done when the force in horizontal direction
W = mgh = 100 × 1.2 = 120 J ?
13. m = 500 kg, s = 25m, u = 72km/h= 20 m/s, v = 0
(-a) =
S 2
u v
2 2
?
? a =
50
400
= 8m/sec
2
Frictional force f = ma = 500 × 8 = 4000 N
14. m = 500 kg, u = 0, v = 72 km/h = 20m/s
a =
s 2
u v
2 2
?
=
50
400
= 8m/sec
2
force needed to accelerate the car F = ma = 500 × 8 = 4000 N
15. Given, v = a x (uniformly accelerated motion)
displacement s = d – 0 = d
putting x = 0, v
1
= 0
putting x = d, v
2
= a d
a =
s 2
u v
2
2
2
2
?
=
d 2
d a
2
=
2
a
2
force f = ma =
2
ma
2
work done w = FS cos ? = d
2
ma
2
? =
2
d ma
2
?
16. a) m = 2kg, ? = 37°, F = 20 N
From the free body diagram
F = (2g sin ?) + ma ? a = (20 – 20 sin ?)/s = 4m/sec
2
S = ut + ½ at
2
(u = 0, t = 1s, a = 1.66)
= 2m
So, work, done w = Fs = 20 × 2 = 40 J
b) If W = 40 J
S =
F
W
=
20
40
h = 2 sin 37° = 1.2 m
So, work done W = –mgh = – 20 × 1.2 = –24 J
c) v = u + at = 4 × 10 = 40 m/sec
So, K.E. = ½ mv
2
= ½ × 2 × 16 = 16 J
17. m = 2kg, ? = 37°, F = 20 N, a = 10 m/sec
2
a) t = 1sec
So, s= ut + ½ at
2
= 5m
37°
?
A ?
A ?
A ?
B ?
v=0
?
v=20 m/s
m=500 kg ?
–a
?
25m
?
a
?
R
mg
?
ma f
?
500 kg ?
a
?
25m
?
R
mg
?
F
F
?
ma
?
ma ?
2g cos ? ?
20N
?
R
?
ma 2gsin ? ?
20N ?
ma ?
mg cos ? ?
20N
?
R
?
mg sin ? ?
?R ?
h
?
37°
?
5m ?
C ?
A ?
B ?
37°
?
Chapter 8
8.4
Work done by the applied force w = FS cos 0° = 20 × 5 = 100 J
b) BC (h) = 5 sin 37° = 3m
So, work done by the weight W = mgh = 2 × 10 × 3 = 60 J
c) So, frictional force f = mg sin ?
work done by the frictional forces w = fs cos0° = (mg sin ?) s = 20 × 0.60 × 5 = 60 J ?
18. Given, m = 25o g = 0.250kg,
u = 40 cm/sec = 0.4m/sec
? = 0.1, v=0
Here, ? R = ma {where, a = deceleration}
a =
m
R ?
=
m
mg ?
= ?g = 0.1 × 9.8 = 0.98 m/sec
2
S =
a 2
u v
2 2
?
= 0.082m = 8.2 cm
Again, work done against friction is given by
– w = ? RS cos ?
= 0.1 × 2.5 × 0.082 × 1 ( ? = 0°) = 0.02 J
? W = – 0.02 J
?
19. h = 50m, m = 1.8 × 10
5
kg/hr, P = 100 watt,
P.E. = mgh = 1.8 × 10
5
× 9.8 × 50 = 882 × 10
5
J/hr
Because, half the potential energy is converted into electricity,
Electrical energy ½ P.E. = 441 × 10
5
J/hr
So, power in watt (J/sec) is given by =
3600
10 441
5
?
? number of 100 W lamps, that can be lit
100 3600
10 441
5
?
?
= 122.5 ?122
20. m = 6kg, h = 2m
P.E. at a height ‘2m’ = mgh = 6 × (9.8) × 2 = 117.6 J
P.E. at floor = 0
Loss in P.E. = 117.6 – 0 = 117. 6 J ? 118 J
21. h = 40m, u = 50 m/sec
Let the speed be ‘v’ when it strikes the ground.
Applying law of conservation of energy
mgh + ½ mu
2
= ½ mv
2
? 10 × 40 + (1/2) × 2500 = ½ v
2
? v
2
= 3300 ? v = 57.4 m/sec ?58 m/sec
22. t = 1 min 57.56 sec = 11.56 sec, p= 400 W, s =200 m
p =
t
w
, Work w = pt = 460 × 117.56 J
Again, W = FS =
200
56 . 117 460 ?
= 270.3 N ? 270 N
23. S = 100 m, t = 10.54 sec, m = 50 kg
The motion can be assumed to be uniform because the time taken for acceleration is
minimum.
Page 5
8.1
SOLUTIONS TO CONCEPTS
CHAPTER – 8
1. M = m
c
+ m
b
= 90kg
u = 6 km/h = 1.666 m/sec
v = 12 km/h = 3.333 m/sec
Increase in K.E. = ½ Mv
2
– ½ Mu
2
= ½ 90 × (3.333)
2
– ½ × 90 × (1.66)
2
= 494.5 – 124.6 = 374.8 ? 375 J
2. m
b
= 2 kg.
u = 10 m/sec
a = 3 m/aec
2
t = 5 sec
v = u + at = 10 + 3 I 5 = 25 m/sec.
?F.K.E = ½ mv
2
= ½ × 2 × 625 = 625 J.
3. F = 100 N
S = 4m, ? = 0°
? = S . F
? ?
???100 × 4 = 400 J ?
4. m = 5 kg
? = 30°
S = 10 m
F = mg
So, work done by the force of gravity
? = mgh = 5 × 9.8 × 5 = 245 J ?
5. F= 2.50N, S = 2.5m, m =15g = 0.015kg.
So, w = F × S ? a =
m
F
=
015 . 0
5 . 2
=
3
500
m/s
2
=F × S cos 0° (acting along the same line)
= 2.5 × 2.5 = 6.25J
Let the velocity of the body at b = U. Applying work-energy principle ½ mv
2
– 0 = 6.25
? V =
015 . 0
2 25 . 6 ?
= 28.86 m/sec.
So, time taken to travel from A to B.
? t =
a
u v ?
=
500
3 86 . 28 ?
? Average power =
t
W
=
3 ) 86 . 28 (
500 25 . 6
?
?
= 36.1
6. Given
j
ˆ
3 i
ˆ
2 r
1
? ?
?
j
ˆ
2 i
ˆ
3 r
2
? ?
So, displacement vector is given by,
2 1
r r r
? ? ?
? ? ? j
ˆ
i
ˆ
) j
ˆ
3 i
ˆ
2 ( ) j
ˆ
2 i
ˆ
3 ( r ? ? ? ? ? ?
?
u=1.66 m/s ?
90kg ? 90kg ?
v=3.33 m/s ?
u=10 m/s ?
2 kg ? a
?
= 3m/s
2
F ?
4m ?
R ?
100 N ?
mg ?
30°
?
5 ?
mg ?
F ?
5 log ?
10m ?
30°
?
v ? ?
A ? B ?
Chapter 8
8.2
So, work done = s F
?
?
? = 5 × 1 + 5(-1) = 0
7. m
b
= 2kg, s = 40m, a = 0.5m/sec
2
So, force applied by the man on the box
F = m
b
a = 2 × (0.5) = 1 N
? = FS = 1 × 40 = 40 J ?
8. Given that F= a + bx
Where a and b are constants.
So, work done by this force during this force during the displacement x = 0 and x = d is given
by
W =
? ?
? ?
d
0
d
0
dx ) bx a ( dx F = ax + (bx
2
/2) = [a + ½ bd] d
9. m
b
= 250g = .250 kg
? = 37°, S = 1m.
Frictional force f = ?R
mg sin ? = ? R ..(1)
mg cos ?? ..(2)
so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J
10. a =
) m M ( 2
F
?
(given)
a) from fig (1)
ma = ?
k
R
1
and R
1
= mg
? ? =
1
R
ma
=
g ) m M ( 2
F
?
b) Frictional force acting on the smaller block f = ?R =
) m M ( 2
F m
mg
g ) m M ( 2
F
?
?
? ?
?
c) Work done w = fs s = d
w = d
) m M ( 2
mF
?
?
=
) m M ( 2
mFd
?
?
11. Weight = 2000 N, S = 20m, ? = 0.2
a) R + Psin ? - 2000 = 0 ..(1)
P cos ? - 0.2 R =0 ..(2)
From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0
P =
? ? ? sin 2 . 0 cos
400
..(3)
So, work done by the person, W = PS cos ? =
? ? ?
?
sin 2 . 0 cos
cos 8000
=
? ? sin 2 . 0 1
8000
=
? ? tan 5
40000
b) For minimum magnitude of force from equn(1)
d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2
putting the value in equn (3)
W =
? ? tan 5
40000
=
) 2 . 5 (
40000
= 7690 J
12. w = 100 N, ? = 37°, s = 2m
R ?
m b g ?
m b a ? F ?
R ?
?R ?
1 m ?
mg ?
37° ?
M
?
F ?
m
?
R 1 ?
? k R 1 ?
ma ?
mg ?
?R 1 ?
R 2 ?
f
?
ma ?
mg ?
?R 2 ?
? ?
R
?
P
?
0.2R ?
2000 N ?
Chapter 8
8.3
Force F= mg sin 37° = 100 × 0.60 = 60 N
So, work done, when the force is parallel to incline.
w = Fs cos ? = 60 × 2 × cos ? = 120 J
In ?ABC AB= 2m
CB = 37°
so, h = C = 1m
?work done when the force in horizontal direction
W = mgh = 100 × 1.2 = 120 J ?
13. m = 500 kg, s = 25m, u = 72km/h= 20 m/s, v = 0
(-a) =
S 2
u v
2 2
?
? a =
50
400
= 8m/sec
2
Frictional force f = ma = 500 × 8 = 4000 N
14. m = 500 kg, u = 0, v = 72 km/h = 20m/s
a =
s 2
u v
2 2
?
=
50
400
= 8m/sec
2
force needed to accelerate the car F = ma = 500 × 8 = 4000 N
15. Given, v = a x (uniformly accelerated motion)
displacement s = d – 0 = d
putting x = 0, v
1
= 0
putting x = d, v
2
= a d
a =
s 2
u v
2
2
2
2
?
=
d 2
d a
2
=
2
a
2
force f = ma =
2
ma
2
work done w = FS cos ? = d
2
ma
2
? =
2
d ma
2
?
16. a) m = 2kg, ? = 37°, F = 20 N
From the free body diagram
F = (2g sin ?) + ma ? a = (20 – 20 sin ?)/s = 4m/sec
2
S = ut + ½ at
2
(u = 0, t = 1s, a = 1.66)
= 2m
So, work, done w = Fs = 20 × 2 = 40 J
b) If W = 40 J
S =
F
W
=
20
40
h = 2 sin 37° = 1.2 m
So, work done W = –mgh = – 20 × 1.2 = –24 J
c) v = u + at = 4 × 10 = 40 m/sec
So, K.E. = ½ mv
2
= ½ × 2 × 16 = 16 J
17. m = 2kg, ? = 37°, F = 20 N, a = 10 m/sec
2
a) t = 1sec
So, s= ut + ½ at
2
= 5m
37°
?
A ?
A ?
A ?
B ?
v=0
?
v=20 m/s
m=500 kg ?
–a
?
25m
?
a
?
R
mg
?
ma f
?
500 kg ?
a
?
25m
?
R
mg
?
F
F
?
ma
?
ma ?
2g cos ? ?
20N
?
R
?
ma 2gsin ? ?
20N ?
ma ?
mg cos ? ?
20N
?
R
?
mg sin ? ?
?R ?
h
?
37°
?
5m ?
C ?
A ?
B ?
37°
?
Chapter 8
8.4
Work done by the applied force w = FS cos 0° = 20 × 5 = 100 J
b) BC (h) = 5 sin 37° = 3m
So, work done by the weight W = mgh = 2 × 10 × 3 = 60 J
c) So, frictional force f = mg sin ?
work done by the frictional forces w = fs cos0° = (mg sin ?) s = 20 × 0.60 × 5 = 60 J ?
18. Given, m = 25o g = 0.250kg,
u = 40 cm/sec = 0.4m/sec
? = 0.1, v=0
Here, ? R = ma {where, a = deceleration}
a =
m
R ?
=
m
mg ?
= ?g = 0.1 × 9.8 = 0.98 m/sec
2
S =
a 2
u v
2 2
?
= 0.082m = 8.2 cm
Again, work done against friction is given by
– w = ? RS cos ?
= 0.1 × 2.5 × 0.082 × 1 ( ? = 0°) = 0.02 J
? W = – 0.02 J
?
19. h = 50m, m = 1.8 × 10
5
kg/hr, P = 100 watt,
P.E. = mgh = 1.8 × 10
5
× 9.8 × 50 = 882 × 10
5
J/hr
Because, half the potential energy is converted into electricity,
Electrical energy ½ P.E. = 441 × 10
5
J/hr
So, power in watt (J/sec) is given by =
3600
10 441
5
?
? number of 100 W lamps, that can be lit
100 3600
10 441
5
?
?
= 122.5 ?122
20. m = 6kg, h = 2m
P.E. at a height ‘2m’ = mgh = 6 × (9.8) × 2 = 117.6 J
P.E. at floor = 0
Loss in P.E. = 117.6 – 0 = 117. 6 J ? 118 J
21. h = 40m, u = 50 m/sec
Let the speed be ‘v’ when it strikes the ground.
Applying law of conservation of energy
mgh + ½ mu
2
= ½ mv
2
? 10 × 40 + (1/2) × 2500 = ½ v
2
? v
2
= 3300 ? v = 57.4 m/sec ?58 m/sec
22. t = 1 min 57.56 sec = 11.56 sec, p= 400 W, s =200 m
p =
t
w
, Work w = pt = 460 × 117.56 J
Again, W = FS =
200
56 . 117 460 ?
= 270.3 N ? 270 N
23. S = 100 m, t = 10.54 sec, m = 50 kg
The motion can be assumed to be uniform because the time taken for acceleration is
minimum.
Chapter 8
8.5
a) Speed v = S/t = 9.487 e/s
So, K.E. = ½ mv
2
= 2250 J
b) Weight = mg = 490 J
given R = mg /10 = 49 J
so, work done against resistance W
F
= – RS = – 49 × 100 = – 4900 J
c) To maintain her uniform speed, she has to exert 4900 j of energy to over come friction
P =
t
W
= 4900 / 10.54 = 465 W
24. h = 10 m
flow rate = (m/t) = 30 kg/min = 0.5 kg/sec
power P =
t
mgh
= (0.5) × 9.8 × 10 = 49 W
So, horse power (h.p) P/746 = 49/746 = 6.6 × 10
–2
hp
25. m = 200g = 0.2kg, h = 150cm = 1.5m, v = 3m/sec, t = 1 sec
Total work done = ½ mv
2
+ mgh = (1/2) × (0.2) ×9 + (0.2) × (9.8) × (1.5) = 3.84 J
h.p. used =
746
84 . 3
= 5.14 × 10
–3
26. m = 200 kg, s = 12m, t = 1 min = 60 sec
So, work W = F cos ? = mgs cos0° [ ? = 0°, for minimum work]
= 2000 × 10 × 12 = 240000 J
So, power p =
t
W
=
60
240000
= 4000 watt
h.p =
746
4000
= 5.3 hp. ?
27. The specification given by the company are
U = 0, m = 95 kg, P
m
= 3.5 hp
V
m
= 60 km/h = 50/3 m/sec t
m
= 5 sec
So, the maximum acceleration that can be produced is given by,
a =
5
0 ) 3 / 50 ( ?
=
3
10
So, the driving force is given by
F = ma = 95 ×
3
10
=
3
950
N
So, the velocity that can be attained by maximum h.p. white supplying
3
950
will be
v =
F
p
? v =
950
5 746 5 . 3 ? ?
= 8.2 m/sec.
Because, the scooter can reach a maximum of 8.s m/sec while producing a force of 950/3 N,
the specifications given are some what over claimed.
28. Given m = 30kg, v = 40 cm/sec = 0.4 m/sec s = 2m
From the free body diagram, the force given by the chain is,
F = (ma – mg) = m(a – g) [where a = acceleration of the block]
a =
2s
u2) (v2
=
4 . 0
16 . 0
= 0.04 m/sec
2
mg
?
F
?
ma
?
mg
?
F
?
Read MoreFAQs on HC Verma Solutions: Chapter 8 - Work & Energy
| 1. How do I tell the difference between positive and negative work in physics problems? |  |
Ans. Work is positive when force and displacement move in the same direction, and negative when they oppose each other. If you push a box forward and it moves forward, work is positive. If friction acts opposite to motion, that's negative work. The sign depends entirely on the angle between force and displacement vectors, not on whether the object speeds up or slows down.
| 2. Why do we use the work-energy theorem instead of just using Newton's laws? | |
Ans. The work-energy theorem directly connects total work done to change in kinetic energy, making complex motion problems simpler without needing acceleration values. While Newton's laws require force analysis at each instant, this theorem uses energy changes across the entire path. For variable forces or curved trajectories, the energy method bypasses lengthy force calculations and saves time during NEET problem-solving.
| 3. What's the difference between kinetic energy and potential energy, and when do I use each? | |
Ans. Kinetic energy depends on motion (½mv²), while potential energy depends on position or configuration (mgh for gravity). Use kinetic energy formulas when analysing moving objects; apply potential energy when dealing with heights, springs, or stored energy. In conservation of mechanical energy problems, both forms interconvert-as an object falls, potential energy converts to kinetic energy, keeping total mechanical energy constant.
| 4. Can mechanical energy ever be conserved when friction is present? | |
Ans. No, mechanical energy decreases when friction acts because friction converts kinetic and potential energy into heat. However, total energy (mechanical + thermal) remains conserved. In NEET problems, always account for work done by friction separately using the work-energy theorem. If friction is ignored, you can apply conservation of mechanical energy; if present, use work-energy relationships to find final speeds or heights accurately.
| 5. How do I approach power and energy rate problems without getting confused? | |
Ans. Power measures how quickly work is done (P = W/t or P = F·v), while energy represents total capacity to do work. Calculate instantaneous power using force multiplied by velocity; find average power by dividing total work by time interval. For variable power situations in Class 11 physics, break the motion into intervals or use calculus-based approaches. Refer to flashcards and mind maps on EduRev to visualise power-energy relationships clearly.
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